A river has a steady speed of vs. A student swims upstream a distance d and back

Question

A river has a steady speed of vs. A student swims upstream a distance d and back to the starting point.
(a) If the student can swim at a speed of v in still water, how much time tup does it take the student to swim upstream a distance d? Express the answer in terms of d, v, and vs.
tup =

(b) Using the same variables, how much time tdown does it takes to swim back downstream to the starting point?
tdown =

(c) Sum the answers found in parts (a) and (b) and show that the time ta required for the whole trip can be written as
ta =
2d/v
1 - vs2/v2
.
(Do this on paper. Your instructor may ask you to turn in this work.)
(d) How much time tb does the trip take in still water? (Use the following as necessary: d and v.)
tb =

(e) Which is larger, ta or tb?
ta
tb

Is it always larger?
Yes
No

Explanation / Answer

SOLUTION: River has a steady speed of ,vs Width of the river,d Student can swim at a speed of, v (a) In order to swim up stream, the student must swim against the river's current. so relative speedof the student must be , v-vs thus, requireed time is, tup= d/v-vs _________________________________________________________________ _________________________________________________________________ (b) In order to swim downstream, the student must swim in the river's current direction. so relative speedof the student must be , v+vs thus, requireed time is, tup= d/v+vs _________________________________________________________________ _________________________________________________________________ (c) time required for total trip ta=(d/v-vs) +(d/v+vs) =2dv/v2-vs2 ____________________________________________________ ____________________________________________________ (d) the trip in still water takes, tb=2d/v ___________________________________________________ ___________________________________________________ (e) It is clear that, 2d/v >2dv/v2-vs2 Hence, tb>ta No, it happens only in the direction against to ' river flow

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